Question

The equation of a circle is ​x2+4x+y2−10y+13=0​ .

Answer 1

X^2+4x+4+y^2-10y+25=-13+25+4

(x+2)^2 + (y-5)^2= 16
Third option

Answer 2

Answer:

$(x+2)^2+(y-5)^2=16$

Step-by-step explanation:

Given : $x^2+4x+y^2-10y+13=0$

To Find: What is the equation of the circle in standard form?

Solution:

$x^2+4x+y^2-10y+13=0$

$x^2+4x+2^2-2^2+y^2-10y+5^2-5^2+13=0$

$(x+2)^2-4+(y-5)^2-25+13=0$

$(x+2)^2+(y-5)^2-29+13=0$

$(x+2)^2+(y-5)^2-16=0$

$(x+2)^2+(y-5)^2=16$

Hence the equation of the circle in standard form is $(x+2)^2+(y-5)^2=16$

So, Option C is correct.